# 5A constant current/ constant voltage using LM350

I take the circuit below given in a TI application note, and I have some questions about the way the circuit works:

The circuit is designed as a 5A constant current/constant voltage regulator. First I have examined the voltage loop, I found it to be composed of R6 and R8. This voltage divider sets the output voltage for the circuit.

When I come to the current loop, many questions arised:

• I understand that the PNP transistor used in paralell with the internal pass element of LM350 to share the output current since LM350 cannot supply 5A. But how does the circuit work in the current loop configuration, I did not get it because:
• To control the output current, we usually measure it by means of a sense resistor (shunt resistor). R3 may play that role, but I think only the current supplied by PNP transistor will be monitored by that.
• If we assume the OPAMP LM301A is used to monitor the current, how does it regulate the current? What is the purpose to put it in an integrator form by placing C5 between its inverting input and its output!
• What is the role of D1 and D2?
• Does this answer your question ? electronics.stackexchange.com/questions/473288/… Dec 29, 2019 at 0:16
• @TonyStewartSunnyskyguyEE75 It cannot answer his third question about the integrator. You didn't put a capacitor from (-) to opamp output. (Not that it matters, because the OP misunderstands the purpose of that capacitor, I think.)
– jonk
Dec 29, 2019 at 0:18
• Luxina, these are good questions. I'll +1 for now. I won't have time to go through everything but perhaps some of it. In the meantime, take a look at the datasheet for the LM301A and find the schematic. Look for pin 8. See where it connects. Imagine the meaning of $D_2$ if it is reverse biased. Then try and see what happens if for some reason $D_2$ becomes forward biased, as you examine the LM301A schematic. (You need the schematic to understand the diode.)
– jonk
Dec 29, 2019 at 0:40
• @lux If you like the linear V regulator, you add a linear current limiter using a similar diff. amp. on the current shunt and pulldown the Vin- for voltage set point then you don't need a split supply. Dec 29, 2019 at 1:16
• I still stack in the current loop, because the two loops share R8 and the ADJ pin of the LM350 so I could not analyse the circuit properly Dec 30, 2019 at 0:05

• PNP when Vbe=0.66V the current thru LDO is 0.66V/33 Ohm = 20mA.

• Above this load the PNP bypasses the LDO as a current source
• the LDO pulls up enough to regulate the voltage as the load decides how much current it needs.
• R3 senses PNP current and the voltage rise is amplified thru the inverting input with gain via the pot, R2 with negative feedback to reduce the voltage drive using the LDO Vref

• the voltage reference, Vref is the Vou-Vadj on the LDO used by the Vadj pot to Vin-

• C5 adds a low frequency pole to reduce the 2nd order effects of the feedback loop and loss of phase margin towards loop stability

• since Vref = 1.25 , when that is reduced by the conduction of D3,D1 it increases the LDO output voltage the same way a resistor divider does by sensing the output voltage with a resistor ratio. Except the diodes here work as switches with the high open loop gain behind them regulated by the two pots.

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• Sincerly, I have spent a time trying to understand the current loop but I could not. The OPAMP here plays the role of an error amplifier ().The non inverting input is connected to Vout, so I think that the controller takes Vout as its reference. Why the voltage divider of the current loop is reconnected to Vadj and not to ground ? Dec 29, 2019 at 22:49
• Vadj is a fixed Vref =1.25V independent of Vout but when shunted can raise output voltage. Dec 29, 2019 at 23:42
• Could you please give me the DC transfer function of the current loop ? Jan 2, 2020 at 21:58
• here's a simplified design tinyurl.com/tzl2s57 Jan 14, 2020 at 7:07
• Improved Pot tinyurl.com/uxsk2r3 current adj. not added Jan 14, 2020 at 7:18